A particle is projected horizontally from the top of a tower with a velocity `v_(0)`. If v be its velocity at any instant, then the radius of curvature of the path of the particle at that instant is directely proportional to

Options

A. `v^(3)`
B. `v^(2)`
C. `v`
D. `1//v`

Answer

A

Explanation

As we know
`a_(C)=(v^(2))/R`(centripetal acceleration)
from figure , ` g sin theta=(v^(2))/R`
`rArr g.(v_(0))/v=(v^(2))/R` (since `sin theta_(1)=(v_(0))/v`) `rArr R alphav^(3)`